Duality Principle (Boolean Algebras)

This proof is about Duality Principle in the context of Boolean Algebra. For other uses, see Duality Principle.

Theorem

Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.

Then any theorem in $\struct {S, \vee, \wedge, \neg}$ remains valid if both $\vee$ and $\wedge$ are interchanged, and also $\bot$ and $\top$ are interchanged throughout the whole theorem.

Proof

Let us take the axioms of a Boolean algebra $\struct {S, \wedge, \vee, \neg}$:

\((\text {BA}_1 0)\)	$:$							$S$ is closed under $\vee$, $\wedge$ and $\neg$
\((\text {BA}_1 1)\)	$:$							Both $\vee$ and $\wedge$ are commutative
\((\text {BA}_1 2)\)	$:$							Both $\vee$ and $\wedge$ distribute over the other
\((\text {BA}_1 3)\)	$:$							Both $\vee$ and $\wedge$ have identities $\bot$ and $\top$ respectively
\((\text {BA}_1 4)\)	$:$							$\forall a \in S: a \vee \neg a = \top, a \wedge \neg a = \bot$

It can be seen by inspection that exchanging $\wedge$ and $\vee$, and $\bot$ and $\top$ throughout, does not change the axioms.

Thus, what you get is a Boolean algebra again.

Hence the result.

$\blacksquare$

Sources

• 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $\S 1.5$: Theorem $1.14$
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Boolean algebra
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): duality: 2.
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Boolean algebra
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): duality: 2.